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Double integral of Logarithm

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Post Tue Oct 04, 2016 2:01 pm

Posts: 105
How to obtain the closed form of the following integral
\[\int\limits_0^1 {\int\limits_0^x {\frac{{{{\ln }^k}\left( {1 - t} \right){{\ln }^m}\left( {1+x} \right)}}{{\left( {1 + t} \right)\left( {1 - x} \right)}}dtdx} } = ?\]
where $k$ and $m$ are positve integers.


Post Thu Oct 06, 2016 12:59 pm

Posts: 105

x is Integral variable. \[\int\limits_0^1 {\int\limits_0^x {\frac{{{{\ln }^k}\left( {1 - t} \right){{\ln }^m}\left( {1 + x} \right)}}{{\left( {1 + t} \right)\left( {1 - x} \right)}}dtdx} } = \int\limits_0^1 {\frac{{{{\ln }^m}\left( {1 + x} \right)}}{{1 - x}}\int\limits_0^x {\frac{{{{\ln }^k}\left( {1 - t} \right)}}{{1 + t}}dtdx} } \]


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